Question

chứng minh rằng \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}\)

Answer 1

x,y.z>0 nha

\(\dfrac{xy+yz+zx}{xyz}\ge\dfrac{9}{x+y+z}\)

\(3xyz+\left(x^2y+y^2z+z^2x\right)+\left(x^2z+y^2x+z^2y\right)\ge9xyz\)

Ta cần cm bđt: \(3xyz+\left(x^2y+y^2z+z^2x\right)+\left(x^2z+y^2x+z^2y\right)\ge9xyz\)

Áp dụng BĐT Cô si, ta có:

\(3xyz+\left(x^2y+y^2z+z^2x\right)+\left(x^2z+y^2x+z^2y\right)\ge3xyz+3\sqrt[3]{\left(xyz\right)^3}+3\sqrt[3]{\left(xyz\right)^3}\)

\(3xyz+\left(x^2y+y^2z+z^2x\right)+\left(x^2z+y^2x+z^2y\right)\ge9xyz\)

⇒ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}\)